Question

The heat of combustion of $C_x{H}_y$, carbon and hydrogen are a, b and c cal respectively. The heat of formation of $C_x{H}_y$ will be:

• A. $-(xb+\frac{yc}{2}+\frac{a}{2})$ cal
• B. $-(xb+\frac{yc}{2}-a)$ cal
• C. $(xb-\frac{yc}{2}+\frac{a}{2})$ cal
• D. $(xb-\frac{yc}{2}-\frac{a}{2})$ cal

Answer 1

Answer: (B)

$-(xb+\frac{yc}{2}-a)$ cal

The reaction can be given as: $xC+\frac{y}{2}{H}_2⟶C_x{H}_y$

$\Delta H_f\left(C_x{H}_y\right)=\sum {\Delta }_{C}H\left(products\right)-\sum \Delta {}_{C}H\left(reactants\right)=x{\Delta }_{C}H\left(C\right)+\frac{y}{2}\Delta {}_{C}H\left(H_2\right)-\Delta {}_{C}H\left(C_x{H}_y\right)$

$=(x\times b)+(\frac{y}{2}\times c)-acal=-(xb+\frac{yc}{2}-a)cal$

Therefore, the heat of formation of $C_x{H}_y$ is $-(xb+\frac{yc}{2}-a)cal$